Let `veca=2hati+3hatj+4hatk, vecb=hati-2hatj+jhatk` and `vecc=hati+hatj-hatk.` If `vecr xx veca =vecb` and `vecr.vec c=3,` then the value of `|vecr|` is equal to

Options

A. `sqrt(155)`
B. `sqrt(17)`
C. `2sqrt(17)`
D. 3

Answer

Let's solve the problem step by step.

Given:
\[ \vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}, \quad \vec{b} = \hat{i} - 2\hat{j} + \hat{k}, \quad \vec{c} = \hat{i} + \hat{j} - \hat{k} \]
We know:
\[ \vec{r} \times \vec{a} = \vec{b} \]
\[ \vec{r} \cdot \vec{c} = 3 \]

### Step 1: Calculate \(\vec{a} \cdot \vec{c}\)

To find \(\vec{a} \cdot \vec{c}\):
\[ \vec{a} \cdot \vec{c} = (2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (\hat{i} + \hat{j} - \hat{k}) \]
Calculating the dot product:
\[ = 2 \cdot 1 + 3 \cdot 1 + 4 \cdot (-1) = 2 + 3 - 4 = 1 \]

### Step 2: Use the relation from the cross product

From the relation \(\vec{r} \times \vec{a} = \vec{b}\), we can express it in terms of the scalar triple product:
\[ \vec{r} \times \vec{a} = \vec{b} \implies \vec{r} \cdot \vec{c} \cdot \vec{a} - \vec{a} \cdot \vec{c} \cdot \vec{r} = \vec{b} \times \vec{c} \]

### Step 3: Substitute known values

We know that \(\vec{r} \cdot \vec{c} = 3\) and \(\vec{a} \cdot \vec{c} = 1\):
\[ 3\vec{a} - 1\vec{r} = \vec{b} \times \vec{c} \]

### Step 4: Calculate \(\vec{b} \times \vec{c}\)

Now, we need to calculate \(\vec{b} \times \vec{c}\):
\[ \vec{b} = \hat{i} - 2\hat{j} + \hat{k}, \quad \vec{c} = \hat{i} + \hat{j} - \hat{k} \]
Using the determinant method:
\[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ 1 & 1 & -1 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} -2 & 1 \\ 1 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -2 \\ 1 & 1 \end{vmatrix} \]
Calculating each of the 2x2 determinants:
\[ = \hat{i}((-2)(-1) - (1)(1)) - \hat{j}((1)(-1) - (1)(1)) + \hat{k}((1)(1) - (-2)(1)) \]
\[ = \hat{i}(2 - 1) - \hat{j}(-1 - 1) + \hat{k}(1 + 2) \]
\[ = \hat{i}(1) + \hat{j}(2) + \hat{k}(3) = \hat{i} + 2\hat{j} + 3\hat{k} \]

### Step 5: Substitute back into the equation

Now substituting back:
\[ 3\vec{a} - \vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} \]
Substituting \(\vec{a}\):
\[ 3(2\hat{i} + 3\hat{j} + 4\hat{k}) - \vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} \]
\[ (6\hat{i} + 9\hat{j} + 12\hat{k}) - \vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} \]
Rearranging gives:
\[ \vec{r} = (6\hat{i} + 9\hat{j} + 12\hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) \]
\[ = (6 - 1)\hat{i} + (9 - 2)\hat{j} + (12 - 3)\hat{k} = 5\hat{i} + 7\hat{j} + 9\hat{k} \]

### Step 6: Find the modulus of \(\vec{r}\)

Now, we find \(|\vec{r}|\):
\[ |\vec{r}| = \sqrt{(5)^2 + (7)^2 + (9)^2} = \sqrt{25 + 49 + 81} = \sqrt{155} \]

Thus, the value of \(|\vec{r}|\) is \(\sqrt{155}\).